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If it's a fair coin... (the question doesn't specify).
That's a simple and elegant way to look at it, morningstar. I think that would have convinced me immediately when I first heard this problem. The one that actually reached me first was (as previously mentioned) extending it to 1000 doors and one car, where you pick a door (1/1000) and all the other doors, except the one with the car, are opened. Guess I'm a sucker for reductio ad absurdum![]()
Well if the twist is that it is a double headed coin then that's pretty lame
To be clearer you could have said:That's a simple and elegant way to look at it, morningstar. I think that would have convinced me immediately when I first heard this problem. The one that actually reached me first was (as previously mentioned) extending it to 1000 doors and one car, where you pick a door (1/1000) and all the other doors, except the one with the car, are opened. Guess I'm a sucker for reductio ad absurdum![]()
Here is a fun basic one: How many people must be in a room before the probability that some share a birthday becomes at least 50 percent?
Birth year doesn't matter.
Very close.Without doing the math, I think it's around 20/21 if I remember correctly.
Suppose you pick door #1, but you lie to the host and say door #3.Since we have so many experts on probability here, I thought we could play around with some puzzles. We'll start with an oldie.
"Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?"
Well? Do you stay with your original choice or do you switch? By staying or switching does one give you any better/worse odds?
Thanks!Love it.
Too late.Suppose you pick door #1, but you lie to the host and say door #3.
The host then opens door #1, which has a goat.
Knowing you lied, should you stick with door #3, or switch?
You just won a goat .Suppose you pick door #1, but you lie to the host and say door #3.
The host then opens door #1, which has a goat.
Knowing you lied, should you stick with door #3, or switch?
Someone that understood Boolean Algebra would understand.How?
By sticking or switching?
I suspect that there are more than 262144 people who have flipped coins, thus an isolated series of 18 heads is not unexpected.I have to post this video now (if the new software will let me).
GUILDENSTERN: Heads. He keeps flipping the coin. Heads … Heads … Heads … Heads … Heads … Heads … Heads … Heads … Heads … Heads … Heads … Heads … Heads … Heads … Heads … Heads … Heads … Heads …He throws it up again, and Rosencrantz catches it, looks at the coin and throws it back to Guildenstern. Heads. Rosencrantz gets a coin out of his purse and flips it, covering it with has hand. Bet. Heads, I win. Rosencrantz looks at the coin, says nothing, and throws it to Guildenstern, who covers it with his hand. Again. Guildenstern looks at the coin. Heads. Rosencrantz gets out another coin.
The two continue on horseback, with Guildenstern flipping a coin as he rides. Heads … Heads … Heads … Heads … he drops the coin, and it rolls down the hill. It comes up heads.
GUILDENSTERN: It must be indicative of something besides the redistribution of wealth. He flips a coin to Rosencrantz, who looks at it.
ROSENCRANTZ: Heads.
GUILDENSTERN: A weaker man might be moved to reexamine his faith. If nothing else, at least in the law of probability. He flips another coin to Rosencrantz.
ROSENCRANTZ: Heads.
GUILDENSTERN: Consider. One. Probability is a factor which operates within natural forces. Two. Probability is not operating as a factor. Three. We are now held within un- sub- or super-natural forces. Discuss.
Before the door was opened, the possibilities were:
Code:Door 1. Door 2. Door 3 Car. Goat 1. Goat 2 Car. Goat 2. Goat 1 Goat 1. Car. Goat 2 Goat 1. Goat 2. Car Goat 2. Car. Goat 1 Goat 2. Goat 1. Car
After the door is opened, possibilities 4 and 6 are ruled out. So the remaining possibilities are:
Therefore the conditional probability of a car being behind door 1 is 1/2Code:Door 1. Door 2. Door 3 Car. Goat 1. Goat 2 Car. Goat. 2. Goat 1 Goat 1. Car. Goat 2 Goat 2. Car. Goat 1
Yours Remaining Hosts
Car. Goat 1. Goat 2
Car. Goat 2. Goat 1
Goat 1. Car. Goat 2
Goat 1. Goat 2. Car
Goat 2. Car. Goat 1
Goat 2. Goat 1. Car
Yours Remaining Hosts
Car. Goat 1. Goat 2
Car. Goat 2. Goat 1
Goat 1. Car. Goat 2
Goat 1. Car Goat 2.
Goat 2. Car. Goat 1
Goat 2. Car Goat 1.